Q249 / 270 · NDA 2015 (Phase 2) · Mathematics

Consider the function $f(x)=\left\{\begin{array}{cc} -2 \sin x & \text { if } x \leq-\frac{\pi}{2} \\ A \sin x+B & \text { if }-\frac{\pi}{2} \lt x \lt \frac{\pi}{2} \\ \cos x & \text { if } x \geq \frac{\pi}{2} \end{array}\right.$ which is continuous everywhere. The value of A is

  • A
    1
  • B
    0
  • Correct answer
    C
    -1
  • D
    -2

Solution

$\because f(x)=\left\{\begin{array}{ccc}-2 \sin x & \text { if } & x \leq-\pi / 2 \\ A \sin x+B & \text { if } & -\pi / 2 \lt x \lt \pi / 2 \\ \cos x & \text { if } & x \geq \pi / 2\end{array}\right.$ $\begin{aligned} & \because \mathrm{f}(\mathrm{x}) \text { is continuous every where: } \\ & \therefore \lim _{\mathrm{x} \rightarrow \pi / 2} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow \pi / 2} \mathrm{f}(\mathrm{x}) \\ & \Rightarrow \lim _{\mathrm{x} \rightarrow \pi / 2} \cos \mathrm{x}=\lim _{\mathrm{x} \rightarrow \pi / 2} \mathrm{~A} \sin \mathrm{x}+\mathrm{B} \\ & \Rightarrow \mathrm{~A}+\mathrm{B}=0 \end{aligned}$ Also; $\lim _{x \rightarrow-\pi / 2^{+}} f(x)=\lim _{x \rightarrow-\pi / 2^{-}} f(x)$ $\begin{aligned} & \Rightarrow \lim _{x \rightarrow-\pi / 2} \mathrm{~A} \sin \mathrm{x}+\mathrm{B}=\lim _{\mathrm{x} \rightarrow-\pi / 2}-2 \sin \mathrm{x} \\ & \Rightarrow-\mathrm{A}+\mathrm{B}=2 \end{aligned}$ is from eq(1) \& (2) we get, $A=-1$

Hence, the correct answer is option: 'C'.

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